# An inequality involving weight $|x|^\alpha$

Background:

Let $$u:\mathbb{R}^n\to \mathbb{R}$$. Then the paper considers the following problem

\begin{align*} -\operatorname{div}(w(x) \nabla u) &= w(x) \text{ in } \Omega \\ u &= 0 \text{ on } \partial \Omega\\ u_\nu&= c\text{ on } \partial \Omega \\ \nabla^2\log w(\nabla u,\nabla u)+\frac{(\nabla \log w\cdot \nabla u)^2}{\alpha}&\leq 0 \text{ in }\Omega \end{align*} where $$w$$ is a homogenous weight such that $$\nabla w \cdot x = \alpha x$$ for $$\alpha >0$$ for eg. $$w(x)=|x|^\alpha.$$ Then the authors prove the following theorem:

Theorem 1.1 Let $$Ω$$ be a connected and bounded domain with the smooth boundary $$∂Ω$$ in $$\mathbb{R}^n$$. If there exists a solution $$u ∈ C^2(Ω)$$ of the overdetermined problem above, then $$Ω$$ is a ball centered at 0.

Their strategy to prove this theorem relies on showing that $$u$$ is radial. However, their proof depends crucially on the inequality involving the weights given above. I am trying to verify if the inequality holds in general for $$w(x)=|x|^\alpha.$$ In the case when $$u$$ is assumed to be radial, the inequality is true. However, when proving Theorem 1.1 one cannot assume that $$u$$ is radial (since that is the goal!) and as I have computed below, the inequality is false when $$w(x)=|x|^\alpha.$$

Computation:

First, $$\frac{(\nabla \log w\cdot \nabla u)^2}{\alpha}=\frac{\alpha}{|x|^4}(x\cdot \nabla u)^2$$

since

$$\nabla\log w = \frac{\alpha}{|x|^2} x.$$

I am guessing that (the authors did not define this in the paper)

$$\nabla^2\log w(\nabla u,\nabla u)=\sum_{i,j}D_{ij}\log wD_iuD_ju$$

where $$D_i u$$ is the partial derivative wrt to $$x_i$$ of $$u.$$ This gives me

$$\nabla^2\log w(\nabla u,\nabla u)=\frac{\alpha |\nabla u|^2}{|x|^2}-\frac{2\alpha (x\cdot \nabla u)^2}{|x|^4}$$

since

$$D_{ij}\log w = \frac{\alpha}{|x|^2}\delta_{ij}-\frac{2\alpha}{|x|^4}x_i x_j.$$

Therefore, $$\nabla^2\log w(\nabla u,\nabla u)+\frac{(\nabla \log w\cdot \nabla u)^2}{\alpha}=\frac{\alpha |\nabla u|^2}{|x|^2}-\frac{\alpha (x\cdot \nabla u)^2}{|x|^4}\\ =\frac{\alpha |\nabla u|^2}{|x|^2}\left(1-\cos^2(\theta_x)\right)\geq 0,$$

where $$\theta_x$$ is the angle between $$x$$ and the gradient of $$u$$, however, this contradicts the claim in the paper. Where have I made a mistake?

For your question (or statement) about a particular piece of a paper to be more effective, you should refer, not just to the paper, but to that particular piece of it.

It appears that the piece in question is the paragraph right after Theorem 1.1 in the paper, which begins with the words "When $$u$$ is a radial symmetric solution" and contains a rather detailed derivation of the claimed inequality, which in this case turns out to be the equality.

That derivation suggests that your understanding of the notation $$\nabla^2\log w(\nabla u,\nabla u)$$ (unexplained in the paper) is correct. Your derivation, too, seem correct.

At the very end of it, if you take into account the condition that $$u$$ is radial symmetric, you will get $$\cos^2(\theta_x)=1$$, which will allow you to replace your last inequality $$\ge0$$ by $$=0$$ and hence by $$\le0$$, as desired.

• Thank you for your answer. I will add more details related to the paper in the question. But from what I understand is that the goal is to show that u is radially symmetric since that would imply that the domain is a ball. However, the authors assume that to justify the inequality which is a bit circular, no? Jun 22, 2022 at 13:52
• @Student : I am not sure what pieces of the paper you are talking about. As I said, you should be as specific as possible in everything you refer to. Jun 22, 2022 at 13:58
• @Student : Do you have a further response to my answer and comment? Jun 24, 2022 at 21:21
• I have added more details to my question. Perhaps now it is clear what my question is? Jun 25, 2022 at 9:17